Develop Reference

How to use vlook up in python to find text in a dataframe?

I would like to use something like vlook-up/map function in python.
I have only a portion of entire name of some companies. i would like to know if the company is into the dataset, as the follow example.
Thank you

I can recreate the results checking one list against another. It's not very clear or logical what your match criteria are. "john usa" is a successful match with "aviation john" on the basis that "john" appears in both. But would "john usa" constitute a match with "usa mark sas" since "usa" appears in both? What about hyphens, comma's, etc?
It would help if this was cleared up.
In any case, I hope the following will help, good luck:-
#create two lists of tuples based on the existing dataframes.
check_list = list(df_check.to_records(index=False))
full_list = list(df_full.to_records(index=False))
#create a set - entries in a set are unique
results=set()
for check in check_list: #for each record to check...
for search_word in check[0].split(" "): #take the first column and split it into its words using space as a delimiter
found=any(search_word in rec[0] for rec in full_list) #is the word a substring of any of the records in full list? True or False
results.add((check[0], found)) #add the record we checked to the set with the result (the set avoids duplicate entries)
#build a dataframe based on the results
df_results=df(results, columns=["check", "found"])

df1['in DATASET'] = df1['NAME'].isin(df2['FULL DATASET'])

Trying to create a function to search for a word in a string

it's my first time posting but I have a question regarding trying to create a function in python that will search a list of strings and return any words that I am looking for. Here is what I have so far:
def search_words(data, search_words):
keep = []
for data in data:
if data in search_words:
keep.append(data)
return keep
Here is the data I am searching through and the words I am trying to find:
data = ['SHOP earnings for Q1 are up 5%',
'Subscriptions at SHOP have risen to all-time highs, boosting sales',
"Got a new Mazda, VROOM VROOM Y'ALL",
'I hate getting up at 8am FOR A STUPID ZOOM MEETING',
'TSLA execs hint at a decline in earnings following a capital expansion program']
words = ['earnings', 'sales']
Upon doing print(search_words(data=data, search_words=words)) my list (keep) returns empty brackets [ ] and I am unsure of how to fix the issue. I know that searching for a word in a string is different than looking for a number in a list but I cannot figure out to modify my code to account for that. Any help would be appreciated.

You can use the following. This will keep all the sentences in data that contain at least one of the words:
keep = [s for s in data if any(w in s for w in words)]

Since they are all strings, instead of looping over them all just combine them all and search that. Also make words a set:
[word for word in ' '.join(data).split() if word in words]
Using regex:
re.findall('|'.join(words), ''.join(data))
['earnings', 'sales', 'earnings']

You can use the following. This will keep all the sentences in data that contain at least one or both of the words:
This is a form of programs for beginners.
data, search_words must be a list.
def search_words(data, search_words):
keep = []
for dt in data:
for sw in search_words:
if sw in dt:
keep.append(dt)
return keep

Matching two-word variants with each other if they don't match alphabetically

I'm doing a NLP project with my university, collecting data on words in Icelandic that exist both spelled with an i and with a y (they sound the same in Icelandic fyi) where the variants are both actual words but do not mean the same thing. Examples of this would include leyti (an approximation in time) and leiti (a grassy hill), or kirkja (church) and kyrkja (choke). I have a dataset of 2 million words. I have already collected two wordlists, one of which includes words spelled with a y and one includes the same words spelled with a i (although they don't seem to match up completely, as the y-list is a bit longer, but that's a separate issue). My problem is that I want to end up with pairs of words like leyti - leiti, kyrkja - kirkja, etc. But, as y is much later in the alphabet than i, it's no good just sorting the lists and pairing them up that way. I also tried zipping the lists while checking the first few letters to see if I can find a match but that leaves out all words that have y or i as the first letter. Do you have a suggestion on how I might implement this?

Try something like this:
s = "trydfydfgfay"
l = list(s)
candidateWords = []
for idx, c in enumerate(l):
if c=='y':
newList = l.copy()
newList[idx] = "i"
candidateWord = "".join(newList)
candidateWords.append(candidateWord)
print(candidateWords)
#['tridfydfgfay', 'trydfidfgfay', 'trydfydfgfai']
#look up these words to see if they are real words

I do not think this is a programming challenge, but looks more like an NLP challenge itself. The spelling variations are often a hurdle that one would face during preprocessing.
I would suggest that you use an Edit-distance based approaches to identify word pairs that allow for some variations. Specifically for the kind of problem that you have described above, I would recommend "Jaro Winkler Distance". This method allows to give higher similarity score between word pairs that shows variations between specific character pairs, say y and i.
All these approaches are implemented in Jellyfish library.
You may also have a look at the fuzzywuzzy package as well. Hope this helps.

So this accomplishes my task, kind of an easy not-that-pretty solution I suppose but it works:
wordlist = open("data.txt", "r", encoding='utf-8')
y_words = open("y_wordlist.txt", "w+", encoding='utf-8')
all_words = []
y_words = []
for word in wordlist:
word = word.lower()
all_words.append(word)
for word in all_words:
if "y" in word:
y_words.append(word)
word_dict = {}
for word in y_words:
newwith1y = word.replace("y", "i",1)
newwith2y = word.replace("y", "i",2)
newyback = word[::-1].replace("y", "i",1)
newyback = newyback[::-1]
word_dict[word] = newwith1y
word_dict[word] = newwith2y
word_dict[word] = newyback
for key, value in word_dict.items():
if value in all_words:
y_wordlist.write(key)
y_wordlist.write(" - ")
y_wordlist.write(value)
y_wordlist.write("\n")

Python Text processing (str.contains)

I am using str.contains for text analytics in Pandas. If for the sentence "My latest Data job was an Analyst" , I want a combination of the words "Data" & "Analyst" but at the same time I want to specify the number of words between the two words used for the combination( here it is 2 words between "Data" and "Analyst".Currently I am using (DataFile.XXX.str.contains('job') & DataFile.XXX.str.contains('Analyst') to get the counts for "job Analyst".
How can I Specify the number of words in between the 2 words in the str.contains syntax.
Thanks in advance

You can't. At least, not in a simple or standardized way.
Even the basics, like how you define a "word," are a lot more complex than you probably imagine. Both word parsing and lexical proximity (e.g. "are two words within distance D of one another in sentence s?") is the realm of natural language processing (NLP). NLP and proximity searches are not part of basic Pandas, nor of Python's standard string processing. You could import something like NLTK, the Natural Language Toolkit to solve this problem in a general way, but that's a whole 'nother story.
Let's look at a simple approach. First you need a way to parse a string into words. The following is rough by NLP standards, but will work for simpler cases:
def parse_words(s):
"""
Simple parser to grab English words from string.
CAUTION: A simplistic solution to a hard problem.
Many possibly-important edge- and corner-cases
not handled. Just one example: Hyphenated words.
"""
return re.findall(r"\w+(?:'[st])?", s, re.I)
E.g.:
>>> parse_words("and don't think this day's last moment won't come ")
['and', "don't", 'think', 'this', "day's", 'last', 'moment', "won't", 'come']
Then you need a way to find all the indices in a list where a target word is found:
def list_indices(target, seq):
"""
Return all indices in seq at which the target is found.
"""
indices = []
cursor = 0
while True:
try:
index = seq.index(target, cursor)
except ValueError:
return indices
else:
indices.append(index)
cursor = index + 1
And finally a decision making wrapper:
def words_within(target_words, s, max_distance, case_insensitive=True):
"""
Determine if the two target words are within max_distance positiones of one
another in the string s.
"""
if len(target_words) != 2:
raise ValueError('must provide 2 target words')
# fold case for case insensitivity
if case_insensitive:
s = s.casefold()
target_words = [tw.casefold() for tw in target_words]
# for Python 2, replace `casefold` with `lower`
# parse words and establish their logical positions in the string
words = parse_words(s)
target_indices = [list_indices(t, words) for t in target_words]
# words not present
if not target_indices[0] or not target_indices[1]:
return False
# compute all combinations of distance for the two words
# (there may be more than one occurance of a word in s)
actual_distances = [i2 - i1 for i2 in target_indices[1] for i1 in target_indices[0]]
# answer whether the minimum observed distance is <= our specified threshold
return min(actual_distances) <= max_distance
So then:
>>> s = "and don't think this day's last moment won't come at last"
>>> words_within(["THIS", 'last'], s, 2)
True
>>> words_within(["think", 'moment'], s, 2)
False
The only thing left to do is map that back to Pandas:
df = pd.DataFrame({'desc': [
'My latest Data job was an Analyst',
'some day my prince will come',
'Oh, somewhere over the rainbow bluebirds fly',
"Won't you share a common disaster?",
'job! rainbow! analyst.'
]})
df['ja2'] = df.desc.apply(lambda x: words_within(["job", 'analyst'], x, 2))
df['ja3'] = df.desc.apply(lambda x: words_within(["job", 'analyst'], x, 3))
This is basically how you'd solve the problem. Keep in mind, it's a rough and simplistic solution. Some simply-posed questions are not simply-answered. NLP questions are often among them.

count all occurences of each word from a list that appear in several thousand records in python

I have a list of reviews and a list of words that I am trying to count how many times each word shows in each review. The list of keywords is roughly around 30 and could grow/change. The current population of reviews is roughly 5000 with the review word count ranging from 3 to several hundred words. The number of reviews will definitely grow. Right now the keyword list is static and the number of reviews will not be growing to much so any solution to get the counts of keywords in each review will work, but ideally it will be one where there isn't a major performance issue if the number reviews drastically increase or the keywords change and all the reviews have to be reanalyzed.
I have been reading through different methods on stackoverflow and haven't been able to get any to work. I know you can use skikit learn to get the count of each word, but haven't figured out if there is a way to count a phrase. I have also tried various regex expressions. If the keyword list was all single words, I know I could very easily use skikit learn, a loop or regex, but I am having issues when the keyword has multiple words.
Two links I have tried
Python - Check If Word Is In A String
Phrase matching using regex and Python
the solution here is close, but it doesn't count all occurrences of the same word
How to return the count of words from a list of words that appear in a list of lists?
both the list of keywords and reviews are being pulled from a MySQL DB. All keywords are in lowercase. All text has been made lowercase and all non-alphanumeric except spaces have been stripped from the reviews. My original though was to use skikit learn countvectorizer to count the words, but not knowing how to handle counting a phrase I switched. I am currently attempting with loops and regex, but I am open to any solution
# Example of what I am currently attempting with regex
keywords = ['test','blue sky','grass is green']
reviews = ['this is a test. test should come back twice and not 3 times for testing','this pharse contains test and blue sky and look another test','the grass is green test']
for review in reviews:
for word in keywords:
results = re.findall(r'\bword\b',review) #this returns no results, the variable word is not getting picked up
#--also tried variations of this to no avail
#--tried creating the pattern first and passing it
# pattern = "r'\\b" + word + "\\b'"
# results = re.findall(pattern,review) #this errors with the msg: sre_constants.error: multiple repeat at position 9
#The results would be
review1: test=2; 'blue sky'=0;'grass is green'=0
review2: test=2; 'blue sky'=1;'grass is green'=0
review3: test=1; 'blue sky'=0;'grass is green'=1

I would first do it in brute force rather than overcomplicating it and try to optimize it later.
from collections import defaultdict
keywords = ['test','blue sky','grass is green']
reviews = ['this is a test. test should come back twice and not 3 times for testing','this pharse contains test and blue sky and look another test','the grass is green test']
results = dict()
for i in keywords:
for j in reviews:
results[i] = results.get(i, 0) + j.count(i)
print results
>{'test': 6, 'blue sky': 1, 'grass is green': 1}
it's importont that we query the dict with .get, in case we don't have a key set, we don't want to deal with KeyError exception.
If you want to go the complicated route, you can build your own trie and counter structure to do searches in large text files.
Parsing one terabyte of text and efficiently counting the number of occurrences of each word

None of the options you tried search for the value of word:
results = re.findall(r'\bword\b', review) checks for the word word in the string.
When you try pattern = "r'\\b" + word + "\\b'" you check for the string "r'\b[value of word]\b'.
You can use the first option, but the pattern should be r'\b%s\b' % word. That will search for the value of word.